English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (10x)/(4x^2+8x+9)

Lösung

\int \frac{10 x}{4 x ^{2} + 8 x + 9} d x

\frac{5}{4} ln \frac{4}{5} (x + 1)^{2} + 1 - 5 arctan (\frac{2}{5} (x + 1)) + C

Schritte zur Lösung

\int \frac{10 x}{4 x ^{2} + 8 x + 9} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int \frac{x}{4 x ^{2} + 8 x + 9} d x

Vervollständige das Quadrat

4 x^{2} + 8 x + 9 : 4 (x + 1)^{2} + 5

= 10 \cdot \int \frac{x}{4 ( x + 1 ) ^{2} + 5} d x

Wende U-Substitution an

= 10 \cdot \int \frac{u - 1}{4 u ^{2} + 5} d u

Wende integrale Substitution an

= 10 \cdot \int \frac{5 v - 2}{4 5 ( v ^{2} + 1 )} d v

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \frac{1}{4 5} \cdot \int \frac{5 v - 2}{v ^{2} + 1} d v

Multipliziere aus

\frac{5 v - 2}{v ^{2} + 1} : \frac{5 v}{v ^{2} + 1} - \frac{2}{v ^{2} + 1}

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 10 \cdot \frac{1}{4 5} (\int \frac{5 v}{v ^{2} + 1} d v - \int \frac{2}{v ^{2} + 1} d v)

\int \frac{5 v}{v ^{2} + 1} d v = \frac{5}{2} ln v^{2} + 1

\int \frac{2}{v ^{2} + 1} d v = 2 arctan (v)

= 10 \cdot \frac{1}{4 5} (\frac{5}{2} ln v^{2} + 1 - 2 arctan (v))

Ersetze zurück

= 10 \cdot \frac{1}{4 5} (\frac{5}{2} ln (\frac{2}{5} (x + 1))^{2} + 1 - 2 arctan (\frac{2}{5} (x + 1)))

Vereinfache

10 \cdot \frac{1}{4 5} (\frac{5}{2} ln (\frac{2}{5} (x + 1))^{2} + 1 - 2 arctan (\frac{2}{5} (x + 1))) : \frac{5}{4} ln \frac{4}{5} (x + 1)^{2} + 1 - 5 arctan (\frac{2}{5} (x + 1))

= \frac{5}{4} ln \frac{4}{5} (x + 1)^{2} + 1 - 5 arctan (\frac{2}{5} (x + 1))

Füge eine Konstante zur Lösung hinzu

= \frac{5}{4} ln \frac{4}{5} (x + 1)^{2} + 1 - 5 arctan (\frac{2}{5} (x + 1)) + C

\int sin^{6} (x) cos^{2} (x) d x

\int \frac{1}{2 x 9 - x ^{2}} d x

\int x^{2} sin (β x) d x

\int \frac{sec ^{4} ( 2 x )}{tan ^{9} ( 2 x )} d x

\int (sin (x) + 2 x) d x